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Let $J$ be the $n \times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?

Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.

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  • $\begingroup$ This is true for any non-zero eigenvalue, not just eigenvalue $1$. $\endgroup$ – user593746 Dec 6 '18 at 11:19
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    $\begingroup$ Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+\sum_{k=1}^r {{r}\choose{k}} N^k$. You can show that $\sum_{k=1}^r {{r}\choose{k}} N^k$ is nilpotent of order $n$. $\endgroup$ – Eric Dec 6 '18 at 15:40
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    $\begingroup$ The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$. $\endgroup$ – ancientmathematician Dec 6 '18 at 16:34
  • $\begingroup$ My interest is in 0 characteristic. Thanks for your counter example in positive characteristic. $\endgroup$ – user371231 Dec 6 '18 at 18:46
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Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.

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As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.

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  • $\begingroup$ It is not true. Take $3\times 3$ case: $$J-I=\begin{bmatrix}0 & 1& 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix},\quad J^2-I=\begin{bmatrix}0 & 2& \color{red}{1}\\0 & 0 & 2\\0 & 0 & 0\end{bmatrix}.$$ In what sense are they equal? $\endgroup$ – A.Γ. Dec 6 '18 at 12:50
  • $\begingroup$ Both has same rank as $2.$ $\endgroup$ – neelkanth Dec 6 '18 at 13:03
  • $\begingroup$ @neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ. $\endgroup$ – Eric Dec 6 '18 at 15:47
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    $\begingroup$ It’s standard notation if we work in matrices $\endgroup$ – neelkanth Dec 6 '18 at 16:15
  • $\begingroup$ Can you explain why the ranks are same ? $\endgroup$ – user371231 Dec 6 '18 at 16:18

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